| View previous topic :: View next topic |
| Author |
Message |
smclean23
Really wants to Beat The GMAT!
Joined: 23 Jun 2008
Posts: 123
Thanks given: 0
Thanked 0 times in 0 posts
|
 Posted: Sun Jul 06, 2008 5:21 pm Post subject: Problem Solving |
 |
|
A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11 |
|
| Back to top |
|
 |
|
|
szapiszapo
Just gettin' started!
Joined: 30 Jun 2008
Posts: 28
Thanks given: 0
Thanked 2 times in 2 posts
|
| Posted: Mon Jul 07, 2008 4:55 am Post subject: |
|
|
n1 = half of riders with average weight of 180 pounds
n2 = half of riders with average weight of 215 pounds
n = greatest numbers of riders who can ride safely
n1 = n2
n1 + n2 = n = 2n1 = 2n2
2000 = 180n1 + 215n2 = 395n1 = 395n/2
then n = 4000/395 > 4000/400 => n is just over 10, which is the nearest integer
therefore answer is D |
|
| Back to top |
|
|
sethids
Rising GMAT Star
Joined: 23 Jun 2008
Posts: 44
Thanks given: 2
Thanked 2 times in 2 posts
Target GMAT Score: 750
|
| Posted: Mon Jul 07, 2008 8:01 am Post subject: |
|
|
n = Number of Safe riders
2000 = (180*n/2) + (215*n/2)
=> 2000 = 395n/2
=> n = 10 (nearest integer)
Hence D. |
|
| Back to top |
|
|
dalwow
Rising GMAT Star
Joined: 02 Oct 2007
Posts: 32
Thanks given: 7
Thanked 1 times in 1 posts
|
| Posted: Wed Jul 09, 2008 11:59 am Post subject: |
|
|
I found it easier just to use a thought process on this one. We want the most people possible at the lesser weight. So, the average of 215 lbs could be just one person. So, subtract 2000 - 215 and get 1785. Then divide 1785/180 = 9. 9+1=10.
_________________
Still Learning |
|
| Back to top |
|
|
Stuart Kovinsky
GMAT Instructor
Joined: 08 Jan 2008
Posts: 1057
Thanks given: 0
Thanked 151 times in 139 posts
Location: Toronto
GMAT Score: 800
|
| Posted: Wed Jul 09, 2008 1:57 pm Post subject: Re: Problem Solving |
|
|
| smclean23 wrote: | A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11 |
We can eliminate 3 answers doing almost 0 work, with a bit of common sense.
Since the question talks about the average weight of half the riders, we know that the total number of riders must be divisible by 2: eliminate (a), (c) and (e). If you're stuck, you have a 50/50 shot.
To actually solve, common sense again comes to the rescue of the algebra-challenged. If the average weight of half the members is 180 and the average weight of the other half is 215, that means that if we look at 1 person from each group their total weight is 395 lbs - let's round that up to 400.
The total limit is 2000 lbs. 2000/400 = 5. So, we can fit 5 pairs of riders on the elevator, for a max total of 10 people: choose (d).
As a side note, :
| Quote: | | I found it easier just to use a thought process on this one. We want the most people possible at the lesser weight. So, the average of 215 lbs could be just one person. So, subtract 2000 - 215 and get 1785. Then divide 1785/180 = 9. 9+1=10. |
is NOT a safe solution, since the question clearly states that half of the riders belong to each group. If the weight limit of the elevator had been, for example, 2100 lbs, (d) would still be the correct answer, but your method would have given you (e) 11 instead.
_________________
Stuart Kovinsky, B.A. LL.B.
Academic Co-ordinator
Kaplan Test Prep & Admissions
Toronto Office
1-800-KAP-TEST
Learn more about me |
|
| Back to top |
|
|
|
|
dalwow
Rising GMAT Star
Joined: 02 Oct 2007
Posts: 32
Thanks given: 7
Thanked 1 times in 1 posts
|
| Posted: Thu Jul 10, 2008 6:07 am Post subject: |
|
|
Thanks Stuart for pointing that out. I obviously wasn't reading it for clarity.
_________________
Still Learning |
|
| Back to top |
|
|
ildude02
GMAT Destroyer!
Joined: 13 Jan 2008
Posts: 307
Thanks given: 0
Thanked 6 times in 6 posts
|
| Posted: Thu Jul 10, 2008 9:13 am Post subject: Re: Problem Solving |
|
|
| Stuart Kovinsky wrote: | | smclean23 wrote: | A certain elevator has a safe weight limit of 2,000 pounds. What is the greatest possible number of people who can safely ride on the elevator at one time with the average (arithmetic mean) weight of half the riders being 180 pounds and the average weight of the others being 215 pounds?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11 |
We can eliminate 3 answers doing almost 0 work, with a bit of common sense.
Since the question talks about the average weight of half the riders, we know that the total number of riders must be divisible by 2: eliminate (a), (c) and (e). If you're stuck, you have a 50/50 shot.
To actually solve, common sense again comes to the rescue of the algebra-challenged. If the average weight of half the members is 180 and the average weight of the other half is 215, that means that if we look at 1 person from each group their total weight is 395 lbs - let's round that up to 400.
The total limit is 2000 lbs. 2000/400 = 5. So, we can fit 5 pairs of riders on the elevator, for a max total of 10 people: choose (d).
As a side note, :
| Quote: | | I found it easier just to use a thought process on this one. We want the most people possible at the lesser weight. So, the average of 215 lbs could be just one person. So, subtract 2000 - 215 and get 1785. Then divide 1785/180 = 9. 9+1=10. |
is NOT a safe solution, since the question clearly states that half of the riders belong to each group. If the weight limit of the elevator had been, for example, 2100 lbs, (d) would still be the correct answer, but your method would have given you (e) 11 instead. |
Stuart, great strategies to solve them without actually following the conventional way of solving the question. I will try ot remember them, just incase I get stuck with the conventional approach, I know these techniques can atleast help me narrowing the choices. |
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You cannot download files in this forum
|
|
"GMAT" and other GMAC™ trademarks are registered trademarks of the Graduate Management Admission Council™. The Graduate Management Admission Council™ does not endorse, nor is it affiliated in any way with the owner or any content on this website. The opinions expressed here are solely those of the author or those of the members of this website. Copyright © 2008 BTG Test Prep, LLC. Powered by phpBB © 2001, 2005 phpBB Group.
|
|
